Fourier Transform
Signal: $ x(t) = 2tjwe^{-t} u(t) $
$ \mathcal{F} = \int_{-\infty}^{+\infty}x(t) e^{-jwt} \, dt = \int_{-\infty}^{+\infty}2tjwe^{-t} u(t) e^{-jwt} \, dt = \int_{0}^{+\infty}2tjw e^{-jwt^2} \, dt $
let $ u = e^{-jwt^2}, du = -2tjwe^{-jwt^2} dt \Rightarrow dt = \frac{du}{-2tjwe^{-jwt^2}} = \frac{du}{{-2tjwu}} $
$ \Rightarrow \mathcal{F} = \int_{1}^{0}\frac{2tjwu}{-2tjwu} \, du = \int_{1}^{0} 1\, du = \left [u] \right ]_1^0 = \left [ e^{-jwt^2} \right ]_0^\infty = -1 $
